+0

# i dont know how to solve can someone help me?

+3
378
11

6x=x^2

x=?

Guest Nov 17, 2014

#11
+92198
+5

Hi Sabi92,

Welcome to the Web2 forum

This is a very strange place to put your question.

Just put it on a new post next time.  :)

I really like that fact that you have had a go at it yourself and that you have shown us your work :)

$$\\a(2b^2-c)+b(c-2b^2)$$

what you need to notice here is that the brackets contain the same things except the signs are different.

I am going to look at the second bracket. I could just as easily chosen the first bracket :)

$$c-2b^2 = -1(-c+2b^2)=-1(2b^2-c)$$

so

$$+b(c-2b^2)=+b*-1(2b^2-c) = -b(2b^2-c)$$

so

$$\\a(2b^2-c)+b(c-2b^2)\\\\ =a(2b^2-c)-b(-c+2b^2)\\\\ =a{(2b^2-c)}-b{(2b^2-c)}\\\\ Now you have a lots of the green stuff minus b lots of the green stuff\\\\ which equals (a-b) lots of the green stuff\\\\ =(a-b)(2b^2-c)\\\\$$

Melody  Jun 7, 2015
Sort:

#1
+92198
+5

6*x=x*x

x must be 6

Melody  Nov 17, 2014
#2
0

no i have these answers x=0 and x=3 but i dont the way how to solve it

Guest Nov 17, 2014
#3
+19207
+5

i dont know how to solve can someone help me

6x=x^2

x=?

$$\small{\text{  \begin{array}{rcl} 6x & = & x^2 \\ x^2 - 6x &=& 0 \\ \underbrace{x}_{=0}(\underbrace{x-6}_{=0}) &=& 0 \\ \end{array}  }}$$

1. x = 0

2. x-6 = 0 | +6

x    = 6

heureka  Nov 17, 2014
#4
0

thank you very much:)!

Guest Nov 17, 2014
#5
+92198
0

Very nice Heureka :)

Melody  Nov 17, 2014
#6
0

oops im sorry i wrote a wrong question the real question is

6x=2x^2

x=?

so how can i solve it?

Guest Nov 17, 2014
#7
+19207
+5

i dont know how to solve can someone help me

6x=2x^2

x=?

$$\small{\text{  \begin{array}{rcl} 6x & = & 2x^2 \\ 2x^2 - 6x &=& 0 \\ \underbrace{x}_{=0}(\underbrace{2x-6}_{=0}) &=& 0 \\ \end{array}  }}$$

1. x = 0

2. 2x-6 = 0 | +6

2x    = 6 | : 2

x    = 6 /2

x    = 3

heureka  Nov 17, 2014
#8
+92198
+5

It is similar to what Heureka did.

$$\\6x=2x^2\\ 3x^2-6x=0\\ 3x(x-2)=0$$

you should be able to finish it :)

Melody  Nov 17, 2014
#9
0

ok thnk you but its 2x(x-3)=0

Guest Nov 17, 2014
#10
+262
0

i need to do a factorization but i dont get the right answer.help please.

a(2b^2-c)+b(c-2b^2)

that's what i did:

2ab^2-ac+bc-2b^3=a(2b^2-c)+b(c-2b^2)

now what?

sabi92  Jun 7, 2015
#11
+92198
+5

Hi Sabi92,

Welcome to the Web2 forum

This is a very strange place to put your question.

Just put it on a new post next time.  :)

I really like that fact that you have had a go at it yourself and that you have shown us your work :)

$$\\a(2b^2-c)+b(c-2b^2)$$

what you need to notice here is that the brackets contain the same things except the signs are different.

I am going to look at the second bracket. I could just as easily chosen the first bracket :)

$$c-2b^2 = -1(-c+2b^2)=-1(2b^2-c)$$

so

$$+b(c-2b^2)=+b*-1(2b^2-c) = -b(2b^2-c)$$

so

$$\\a(2b^2-c)+b(c-2b^2)\\\\ =a(2b^2-c)-b(-c+2b^2)\\\\ =a{(2b^2-c)}-b{(2b^2-c)}\\\\ Now you have a lots of the green stuff minus b lots of the green stuff\\\\ which equals (a-b) lots of the green stuff\\\\ =(a-b)(2b^2-c)\\\\$$

Melody  Jun 7, 2015

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